Theorem sb4bOLD 2476

Description: Obsolete version of sb4b 2475 as of 21-Feb-2024. (Contributed by NM, 27-May-1997.) Revise df-sb 2069. (Revised by Wolf Lammen, 25-Jul-2023.) (New usage is discouraged.) (Proof modification is discouraged.)

Assertion

Ref	Expression
sb4bOLD	⊢ (¬ ∀𝑥 𝑥 = 𝑡 → ([𝑡 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑)))

Proof of Theorem sb4bOLD

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		nfeqf2 2377	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑡 → Ⅎ𝑥 𝑦 = 𝑡)
2		nfnf1 2153	. . . . . . 7 ⊢ Ⅎ𝑥Ⅎ𝑥 𝑦 = 𝑡
3		id 22	. . . . . . 7 ⊢ (Ⅎ𝑥 𝑦 = 𝑡 → Ⅎ𝑥 𝑦 = 𝑡)
4	2, 3	nfan1 2196	. . . . . 6 ⊢ Ⅎ𝑥(Ⅎ𝑥 𝑦 = 𝑡 ∧ 𝑦 = 𝑡)
5		equequ2 2030	. . . . . . . 8 ⊢ (𝑦 = 𝑡 → (𝑥 = 𝑦 ↔ 𝑥 = 𝑡))
6	5	imbi1d 341	. . . . . . 7 ⊢ (𝑦 = 𝑡 → ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑡 → 𝜑)))
7	6	adantl 481	. . . . . 6 ⊢ ((Ⅎ𝑥 𝑦 = 𝑡 ∧ 𝑦 = 𝑡) → ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑡 → 𝜑)))
8	4, 7	albid 2218	. . . . 5 ⊢ ((Ⅎ𝑥 𝑦 = 𝑡 ∧ 𝑦 = 𝑡) → (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑)))
9	8	pm5.74da 800	. . . 4 ⊢ (Ⅎ𝑥 𝑦 = 𝑡 → ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡 → 𝜑))))
10	1, 9	syl 17	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑡 → ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡 → 𝜑))))
11	10	albidv 1924	. 2 ⊢ (¬ ∀𝑥 𝑥 = 𝑡 → (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡 → 𝜑))))
12		df-sb 2069	. 2 ⊢ ([𝑡 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
13		ax6ev 1974	. . . 4 ⊢ ∃𝑦 𝑦 = 𝑡
14	13	a1bi 362	. . 3 ⊢ (∀𝑥(𝑥 = 𝑡 → 𝜑) ↔ (∃𝑦 𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡 → 𝜑)))
15		19.23v 1946	. . 3 ⊢ (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡 → 𝜑)) ↔ (∃𝑦 𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡 → 𝜑)))
16	14, 15	bitr4i 277	. 2 ⊢ (∀𝑥(𝑥 = 𝑡 → 𝜑) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡 → 𝜑)))
17	11, 12, 16	3bitr4g 313	1 ⊢ (¬ ∀𝑥 𝑥 = 𝑡 → ([𝑡 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 395  ∀wal 1537  ∃wex 1783  Ⅎwnf 1787  [wsb 2068

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-10 2139  ax-12 2173  ax-13 2372

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-ex 1784  df-nf 1788  df-sb 2069

This theorem is referenced by: (None)